Probability of rolling at least one 1 in roll of three dice

[burt]

I was asked to show that the probability of at least one 1 appearing in a thow of three dice is \(125:91\).
This is my work:
The probability of rolling a one on each die is \(\frac{1}{6}\).
The probability of rolling all three ones is \(\frac{1}{6}\bullet\frac{1}{6}\bullet\frac{1}{6}=\frac{1}{216}\)
The probability of rolling two ones is \(\frac{1}{6}\bullet\frac{1}{6}\bullet\frac{5}{6}=\frac{5}{216}\)
The probability of rolling only one one is \(\frac{1}{6}\bullet\frac{5}{6}\bullet\frac{5}{6}=\frac{25}{216}\)
Multiply 25 by 5 by 1 to get 125.
\(216-125=91\)
Therefore, the probability of rolling at least one 1 is 125:91

My problem is that I would have added the three probabilities to get \(\frac{31}{216}\) I only multiplied because I saw that would give me my desired result. Also, if multiplication is the correct step, why not multiply the denominator as well?

Can anyone explain?

 

[Romsek]

a much simpler way of approaching it is

[MATH]P[\text{at least one 1 in 3 rolls}] = 1 - P[\text{no 1's in 3 rolls}] = 1 - \left(\dfrac 5 6 \right)^3 = 1 - \dfrac{125}{216} = \dfrac{91}{216}[/MATH]
The problem you encountered is due to the fact that 1 2 and 3 1's rolled are not mutually exclusive events and thus you can't just add their probabilities.

 

[JeffM]

Burt

125:91 IS NOT A PROBABILITY. Those are odds.

A probability is a non-negative number not greater than 1. You would have got at best a half mark for that answer after all your work.

Learn the difference between odds and a probability.

 

[lev888]

I was asked to show that the probability of at least one 1 appearing in a thow of three dice is \(125:91\).
This is my work:
The probability of rolling a one on each die is \(\frac{1}{6}\).
The probability of rolling all three ones is \(\frac{1}{6}\bullet\frac{1}{6}\bullet\frac{1}{6}=\frac{1}{216}\)
The probability of rolling two ones is \(\frac{1}{6}\bullet\frac{1}{6}\bullet\frac{5}{6}=\frac{5}{216}\)
The probability of rolling only one one is \(\frac{1}{6}\bullet\frac{5}{6}\bullet\frac{5}{6}=\frac{25}{216}\)
Multiply 25 by 5 by 1 to get 125.
\(216-125=91\)
Therefore, the probability of rolling at least one 1 is 125:91

My problem is that I would have added the three probabilities to get \(\frac{31}{216}\) I only multiplied because I saw that would give me my desired result. Also, if multiplication is the correct step, why not multiply the denominator as well?

Can anyone explain?

Yes, you should add probabilities. You are not calculating the 2nd and 3rd cases correctly.
Rolling 2 ones: `1/6*1/6*5/6 + 1/6*5/6*1/6 + 5/6*1/6*1/6 = 15/216`
Rolling 1 one: `1/6*5/6*5/6 + 5/6*1/6*5/6 + 5/6*5/6*1/6 = 75/216`

Total with the first case is `91/216`.

 

[burt]

Burt

125:91 IS NOT A PROBABILITY. Those are odds.

A probability is a non-negative number not greater than 1. You would have got at best a half mark for that answer after all your work.

Learn the difference between odds and a probability.

That's true. The probability would be .4212963. But that point is not really helpful in this case.

 

[burt]

a much simpler way of approaching it is

[MATH]P[\text{at least one 1 in 3 rolls}] = 1 - P[\text{no 1's in 3 rolls}] = 1 - \left(\dfrac 5 6 \right)^3 = 1 - \dfrac{125}{216} = \dfrac{91}{216}[/MATH]
The problem you encountered is due to the fact that 1 2 and 3 1's rolled are not mutually exclusive events and thus you can't just add their probabilities.

Can you explain why this works?

The problem you are outlining makes sense.

 

[pka]

Can you explain why this works?
The problem you are outlining makes sense.

If three dice are tossed, there are \(5^3=125\) ways to have all three show a number greater than one; a \(2~,3~,4~,5~,\text{ or }6\).
Tossing three dice that are \(6^3=216\) possible outcomes Thus the probability of no ones is \(\dfrac{125}{216}\).
The probability of at least one one is \(\dfrac{216-125}{216}=\dfrac{91}{216}\).

 

[Romsek]

Can you explain why this works?

The problem you are outlining makes sense.

It's just a basic property of probability.

The probability of anything happening is 1

An event happening, and an event not happening represent a way of partitioning anything happening into two mutually exclusive events.
As they are mutually exclusive we can add their probabilities.

so (a bit of set notation here) P[A] + P[!A] = 1

P[A] = 1 - P[!A]

In the case of the problem clearly getting at least one 1 and getting no 1's are mutually exclusive and cover all possible occurrences.
and thus P[getting at least one 1] = 1 - P[getting no 1's]

 

[Steven G]

That's true. The probability would be .4212963. But that point is not really helpful in this case.

Yes, the point that 91:216 is not a probability is very useful. If that problem was on a test then the answer would be that one can not show that the probability is 91:216. If a student computed that the probability was 91:216 or .4212963 (w/o saying that does not equal 91:216) then they would get no credit. Now if you realized that the problem was stated incorrectly and wanted to find the probability as an exercise then that is fine, however you should have stated that in your post

 

[Steven G]

I was asked to show that the probability of at least one 1 appearing in a thow of three dice is \(125:91\).
This is my work:
The probability of rolling a one on each die is \(\frac{1}{6}\).
The probability of rolling all three ones is \(\frac{1}{6}\bullet\frac{1}{6}\bullet\frac{1}{6}=\frac{1}{216}\)
The probability of rolling two ones is \(\frac{1}{6}\bullet\frac{1}{6}\bullet\frac{5}{6}=\frac{5}{216}\)
The probability of rolling only one one is \(\frac{1}{6}\bullet\frac{5}{6}\bullet\frac{5}{6}=\frac{25}{216}\)
Multiply 25 by 5 by 1 to get 125.
\(216-125=91\)
Therefore, the probability of rolling at least one 1 is 125:91

My problem is that I would have added the three probabilities to get \(\frac{31}{216}\) I only multiplied because I saw that would give me my desired result. Also, if multiplication is the correct step, why not multiply the denominator as well?

Can anyone explain?

Since in multiplication you have to multiply the denominators and since you did not then your answer is wrong--even if you lucked out and got the correct answer. In my opinion, if you get the correct answer for the wrong reason you get no extra credit for getting the right answer